1. **State the problem:** Solve for $x$ in the equation $16(3)^2 = 81(2^{2x})$. 2. **Write down the equation:** $$16 \times 3^2 = 81 \times 2^{2x}$$ 3. **Calculate the powers and simplify:** $$16 \times 9 = 81 \times 2^{2x}$$ $$144 = 81 \times 2^{2x}$$ 4. **Isolate the exponential term:** $$\frac{144}{81} = 2^{2x}$$ 5. **Simplify the fraction:** $$\frac{\cancel{144}^{16} \times 9}{\cancel{81}^{9} \times 9} = \frac{16}{9} = 2^{2x}$$ 6. **Express $\frac{16}{9}$ as powers:** $$\frac{16}{9} = \frac{2^4}{3^2}$$ Since the right side is $2^{2x}$ and the left side has a factor $3^2$ in the denominator, the bases are different. We can take logarithms to solve for $x$. 7. **Take the logarithm base 2 of both sides:** $$\log_2\left(\frac{16}{9}\right) = \log_2\left(2^{2x}\right)$$ 8. **Use logarithm properties:** $$\log_2(16) - \log_2(9) = 2x$$ 9. **Calculate the logarithms:** $$\log_2(16) = 4$$ $$\log_2(9) = \log_2(3^2) = 2 \log_2(3)$$ 10. **Substitute back:** $$4 - 2 \log_2(3) = 2x$$ 11. **Solve for $x$:** $$x = \frac{4 - 2 \log_2(3)}{2} = 2 - \log_2(3)$$ **Final answer:** $$x = 2 - \log_2(3)$$